To find the absolute value of the complex number $9 - 3i$, we use the formula for the modulus of a complex number $z = a + bi$, which is:

$|z| = \sqrt{a^2 + b^2}$

Here:

• $a = 9$ (real part),
• $b = -3$ (imaginary part).

Step-by-Step Solution:

1. Square the real part $a$ and the imaginary part $b$:

$a^2 = 9^2 = 81, \quad b^2 = (-3)^2 = 9$

2. Add the squares together:

$a^2 + b^2 = 81 + 9 = 90$

3. Take the square root of the sum:

$|z| = \sqrt{90}$

4. Simplify $\sqrt{90}$:

$\sqrt{90} = \sqrt{9 \times 10} = \sqrt{9} \cdot \sqrt{10} = 3\sqrt{10}$

Final Answer:

The absolute value of $9 - 3i$ is $3\sqrt{10}$, which matches option C.

Let me know if you'd like me to expand on any part of this solution!

Related Questions:

1. How do you find the absolute value of any complex number?
2. How can you simplify square roots involving products of numbers?
3. What is the general formula for the modulus of $a + bi$?
4. What are some real-world applications of complex numbers?
5. How do you plot a complex number on the complex plane?

Tip:

To simplify square roots, factorize the number into primes and extract squares. For example, $\sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}$.